g = G M / r² = 6.673 x 10^—11 x 1.985 x 10³⁰ / (7 x 10⁸)² = 270.32 m / Sec.                                         In the theme Gravity Force was obtained the time of liberation as follow:
t = v_e / 2 g        r_e  =  v_e t  =  v_e (v_e / 2 g)  =  _ve² / 2 g
v_e = (7 x 10⁸ x 2 x 270.32)^0.5 = 615,186 m / Sec. = velocity of liberation.

Half of the fore velocity is for to nullify the resistance produced by the celestial body (the Sun); this resistance is manifested by the velocity that had the orbital electrons when they worked producing internal energy into the atoms, before these were compressed; and after the compression work as free particles producing external energy. The gravity in the Sun is so strong that not only affects the exterior shells of the atoms, but also some interior ones, as will be seen forward, in such way, that , some of the shells of the atoms are destroy and their orbital electrons get free, producing external energy, manifested as heat, due to the kinetic energy that had the orbital electrons of the disintegrated shells. Here it is pertinent to remember how the orbital electrons move around the nuclei of the atoms (how move as binary systems). In the seven shell the orbital electrons deflect in each vertex, due to the action of the m⊥7 particle, that produce a velocity of deflection v_⊥7 = 69,100 m / Sec.; in the six shell have: m⊥6 and v⊥6 = 93,979 m / Sec.; in . . . . the second shell: m⊥2 and v⊥2 = 826,580 m / Sec.

Our next step is to relate the liberation velocity of the Sun with the deflection velocity of the atoms: v⊥n; here the fore value is an average one, due to the velocity of the orbital electrons liberated of the disintegrated shells.

The energy of the liberation velocity is:                                                                                           K_L = 0.5 m_e v_e²    0.5 x 9.1091 x 10^--31 x 615,186² = 1.723687 x10^--19 joule                                        The fore velocity correspond to a temperature:                                                                                Tn = 2 K_L / 3 k = 2 x 1.723687 x 10^--19 / 3 x 1.38 x 19^--23 = 8,327º K

In accordance with the accepted theories, in the surface of the Sun the temperature is about 5,000^o K and into the Sun can be several millions degree of temperature. The datum of temperature in the surface is easy to prove with direct observations. The temperuture obtained before is bigger than the observed one; with this is concluded that the heat radiated by the surface of the Sun is bigger than the heat that the surface receives from the inner caps of it. But we consider the problem is is not so simple as was say, there are many factprs have not take in account. The data of the interior temperatures, only could be suppose, in order to be in accordance with some accepted theories, that we consider are too far from reality. With our model are simplify so many facts, and because this ought to be study with special interest this theme.

With all have say here, in my protocol begins to understand how the energy is not lost, but transmit from one particle to another one, and with this criterion is get the idea that all the stars can produce energy during infinite time, with out exhaust; maybe they have an evolution process, after which they return to their original condition, and so on. Determine the inherent temperature of the Earth with this model of liberation velocity.                                                                                             v_e =(r g)^0.5 = (6,378,163 x 2 x 9.81)^0.5 = 11,187 m / Ses.                                                                   K_L = 0.5 x 9.1091 x 10^--31 x 11,187² = 5.7 x 10^--23 joule                                                                     T_n = 2 x 5.7 x 10^--23 / 3 x 1.38 x 10^--23 = 2.^o75K

Monterrey, México, April 15, 2000 Manuel de Hoyos Robles

In the table of radii of the atoms of the periodic table are given the radii of all the atoms, except those of the extreme right column. In accordance with the tables of radii and ionization potentials , we have seen that for an atom that has two molecable electrons in the seven shell, one of them, that is the first binary system, gets ionized first; the condition for this is that the radius of the atom that is going to get ionized gets the required conditions for this, in accordance with the formula: 1≈ i r_a (A / K)³. Although the second binary system does not get ionized with the value of (Ii) given in the table; also gets some degree of excitation when the first binary system got ionized; but the second system needs more excitation than the first one; so, remains with its negative particle (its electron).

From all say in the fore paragraph is concluded that the ionization of a binary system is a function of the radius of it. In the tables of ionization the values of ( i) for the atoms of the extreme right column, are those that affect simultaneously the two binary systems of the seven shell; as both system are in equal structural conditions, we deduce that half of the ionization correspond to the first binary system, and the other half to the second binary system. In accordance with this here will be determine the radii of the mentioned atoms, with the following numerical examples:

₂He⁴;    i  =  24.6 eV.;    He; 12.3:    1 = 12.3 r_He² (A / Z)³ = 12.3 r_He² (2 / 4)³ = 1.538 r_He²;                   r_He² =  1 // 1.588  =   0.63        r_He = 0.80 μ

₁₀Ne²⁰;    i = 21.6 eV.      Ne; 10.8:     1 = 10.8 r_Ne² (10 / 20)³ = 1.35 r_Ne² ;    r_Ne² = 0.741;                     r_Ne = 0.861 μ.

₁₈Ar⁴⁰:      i = 15.8 eV.:      Ar; 79:      1 = 7.9 r_Ar² (18 / 40)³ = 0.72 r_Ar²;      r_Ar² = 1.389;                      r_Ar = 1.179 μ.

₃₆Kr⁸⁴:      i = 14 eV:      Kr; 7.0:      1 = 7.0 r_Kr² (36 / 84)³ = 0.551 r_Kr²;      r_Kr² = 1.848;                       r_Kr = 1.347 μ.

In our model we will consider that the atoms of a body are united by the electrons in a similar way than the molecable electrons unite the atoms of a molecule. For understand better the problem let see the behavior of an orbital electron moving around the nucleus of an atom in the seven shell or orbit. In a vertex of such orbit is produced a velocity: v_^7 = 69,100 m / Sec. (due to the propeller particle m_^7, (see forward) that deflects its orbit; this happens in each atom of the body

Since was seen the theme: Velocity of Sound was given the relation that exist between the velocity of sound and the resistance of a material, in accordance with the formula: v_s = (E g / w)^0.5 . Sound moves in the sever orbit electrons. Considering an isometric distribution of atoms, an electron moves along (n) orbits (7 shell) while the sound advances a diameter of atom at velocity (v_s). when the electron has made 2 n half orbits, the sound had advanced a diameter of atom. When the electron has moves 2 n half orbits: the m⊥7 particle transmits to the adjacent electron of the forward atom the sound effect. With a model is consider that an orbital electron of the seven shell, moves (n) orbits while the sound advances one diameter of the atom. After this the effect of sound advance to the forward atom, in which its orbital electron moves also (n) orbits and the sound advance one diameter of atom; and so on. . . In each time of orbit, a propeller particle m_^7 acts with an impulse equal to: v_^7 me for to advance to the electron of the forward atom; an so on during (n) times in the multimolecules that are formed when the sample is affected by a tension or a compression

Will be consider that for the sound advance as was say, it is requires an energy for pass from one atom to the adjacent one. For the electron to deflect in a vertex (7 orbit) it is required a velocity:   v_^7 = 69,100 m / Sec.; the impulse to do this is:    f_v´ =  v_^7 m_e = 69,100 x 9.1091 x 10^—31  =  6.294 x  10^—26 Kg m /Sec.

The fore impulse is produced every time of vertex:                                                                         tv₇  =  8.27 x 10^—17 Sec. is:    f_v   =  f_v´ / t_v = 6.294 x 10^—26 / 8.27 x 10^—17  =  7.611 x 10^—10 Kg m / Sec^.2    =  7.611 x 10^—10 / 9.81  =  7.7584 x     10^---11 Kg.                                                                  Quantity of atoms there are in a area of   1 m²:                                                                               n = 1 / (2r₇)² = 1 / (2 x 1.13763 x 10^—10)² = 1.9317 x 10¹⁹ atoms / m²                                                 Will consider the sound moving in an iron rod.                                                                         Velocity of sound:   v_s = (E g / Δ)² = (2.1 x 10¹⁰ x 9.81 / 7800)^0.5 = 5,106 m / Sec.

When an orbital electrons makes (2 n) half orbits the sound advance a diameter of atom. In the model will give here of structural resistance, this is produced by the action of a propeller particle m⊥7 acting in a molecable electron; this particle jumps from such electron to other one of the adjacent atom, when is affected by an broken impulse. This propeller particle, not only produce attraction between the negative particle and the positive one of a binary system, trying to avoid they fail; but also acts in a synchronized way in the solid bodies, trying to attract the adjacent atoms to be in contact with its own one. Similar thing happens with the atoms of the multimolecules in which flow the current electrons. In the liquid substances this effect is reduced.

When a solid body is affected by some vibrations ,as that of sound, the contact between the adjacent atoms is reduced, also the structural resistance. When a solid fails due to the action of a force, is produced a vibration in that solid body; if the force is enough big is produced a separation between their atoms, and the molecable electrons try to move from their shells to subshells.. The lineal advance of the effect of m⊥7 is (n) times faster than the lineal advance of the sound. This is all right, but the orbital electrons have two orbits in the seven shell, so they have double resistance. Here will consider that sound has the value given before, because the iron works in an elastic way; if it lost its elasticity due to the effort it is affected; also will be lost its sound velocity, due to the same reason

Considering the two half orbits of the seven shell of the electrons affected by sound, we have:           2 v₇ / π = 2 x 308,570 / π = 196,442 m / Sec.                                                           The lineal velocity of the sound in iron is.                                                                                     vs = 5,106 m / Sec.                                                                                                                 Relation between both velocities:                                                                                                    n_Fe  =  (2 v₇ / π) / v_s  =  196,442 / 5106 = 39                                                                                  he deflection fatigue F_> of the orbital electron, expressed in   Kgs / m² is:                                      F_> = f_> n = 7.7584 x 10^—11 x 1.9317 x 10¹⁹  = 1.49869 x 10⁹  Kgs / m² = 149,869 Kgs / cm²             The failing elastic  fatigue of iron rod will be:   F = F_> / n_Fe = 149,869 / 39 = 3,843 Kgs / cm²

For water we have: v_s = 1,450   m / Src.      n_w  =  (2 v₇ / π) / v_s = 196,442 / 1,450 = 135 F_> = 149,869 Kgs / cm² (compression fatigue) Failing fatigue: F = 149,869 / 135 = 1,119 Kgs / cm²         When water is compressed till the fatigue (F), the seven shell electrons jump to a 7 sub-shell . . . (?), or reduces its confined voume (see theme: Model of Refringent Substance-I-)

Monterrey, México; February /2005.; july 2006; March /07 Manuel de Hoyos Robles

Pienso que a los doctores especialistas de Monterrey, excepto el Dr. José Manuel Ramírez Aranda les hace falta ser más humildes y aceptar muchas verdades que les van a permitir curar la enfermedad producida por el virus VIH.

Pienso que los grandes físicos, químicos, biólogos del mundo son mucho más humildes y que muy pronto aceptar muchas verdades que indirectamente van a permitir solucionar este problema.

There is the idea that the conduction of heat is produced by vibration effects: Maybe this is correct in some partial way, from a mathematical point of view; nevertheless I have other ideas for the conduction of heat, that seem more in accordance with the production of the specific heat, and maybe with other heat phenomenon. In our model is considered that heat in form of particles of corpuscles mh propagates along multimolecules, as do the electric current electrons. The particles mh join with the particles  m_^7.  Because this the radius of the active atom grows Δr in each vertex; due to this and to the modulus of elasticity E of the atoms of the plate, in each radial direction of each vertex is produced a fatigue in a length of the magnitude of a multimolecule of:   f_s = Δr E / l_,m every time of orbit. But this problem will be study in other theme.

I in a metallic plate is produced heat in one side of it; logically it will flows to the opposite side of it; bit at the same time, this heat will try to flow parallel to the side were in which is produced such heat. The model will give here, consider both kinds of flow, not in the instant they initiate, but when they get regulated flow. If we want to know the temperature of a gas, or of a liquid, or of any substance, we can determine them considering a big quantity of atoms or molecules of them. If it is wanted to determine such temperature taking in account only one atom or molecule of the substance will obtain wrong results; much lower or much higher temperatures than the average ones given by many atoms or molecules. This can be proved studding the Brownian movement , in which some particles with determine mass move at much higher kinetic energy than the corresponding to the temperature of the liquid. In our model of electric current it is considered that the last atoms of the multimolecules get much higher temperature than the rest ones of such multimolecule, and also those adjacent to such atom. Will consider that the conduction heat is distributed in the atoms of a body in form of a particle of corpuscles mh, as those of the propeller ones m_^7; these particles are into the orbital electrons (seven shell); the electron moves around the atom during a time of jump and at the same time transmits the heat particle to the adjacent atom of a multimolecule, as will be explained better in forward paragraphs

For to explain the conduction of heat in a body it is required to know how it enters to the body; what part is absorbed by the body; and what part flows over it and afterward leaves the body in a radiation or convection way. Here will be made a model trying to explain all these processes in the most simple and harmonic way . The conduction of heat in a metallic plate from one hotter side to the opposite one, result more illustrative in the in the model we will give here. We consider that all the atoms that are in the hotter side of the plate absorb the heat is produced in such side. Here will take a small surface A of the plate. In figure (1) is show A₁ (continuous line); A₂ (pointed line). A = A₁ + A₂; This surface seems as a rectangle; really are two spiral areas with a thick equal to L_m. In the lines (continue, pointed one), every distance Lm is indicated by a point; in the vertices, also by a number. A = (15 + 1) L_m (7 + 1) L_m = 112 L_m². It was say that the spiral lines are in the center of the spiral area; but are also in the center of a spiral volume that has a thick equal to _Lm. The transversal section f this volume is equal to L_m² atoms; each atom in the transversal section correspond to a multimolecule, in which flow and are absorbed the heat particles.

Has been consider that in the huge celestial bodies, that their interior zones increment their temperature due to the great pressure. In the stars this heat try to move toward the surface, by a conduction effect, following radial directions. In this condition our model gets some complications that will difficult our explanations; so, will be choose a body in which the transmitting of heat will be practically in a perpendicular direction, as in the case of a metallic plate. If the plate is heated in a uniform way in one side, the heat will tries to move to the opposite one, practically in a perpendicular direction to the surface of the plate. If the plate is heated in only one zone of its two surfaces, the heat will try to move also parallel to its surface. Here will be given a model of conduction of heat considering the advance movement of it in one direction, also taking in account the capacity of each electron of each atom of the plate to absorb and transmits heat to the adjacent electron of the adjacent atom

The hat absorbed by the higher temperature side of the plate will move toward the lower temperature will be consider an extensive and flat plate with a uniform thick. In all this process, the absorbed side of the plate; and afterward a part of it will be emitted by this side as a radiation or a convection fluid. Here energy by one side of the plate ought to be equal. to the conduction, the absorbed and the convection and radiated energy by the atoms of the plate. In the fore theme: Specific Heat (II) was seen that the atoms of a body or any substance absorb the half of the heat they receive from other atoms or from the exterior medium: It was say that the seven orbit electrons of the atoms transmit the heat in form of particles of corpuscles mh to the adjacent atoms, in their vertices. A particle mh, formed by corpuscles, can produce heat. The value of mh is obtained by the difference of heat between the two sides of the plate-. The interior shell particles also can absorb the heat is introduced into them, and also can yield some excess heat.

The conduction of heat can be produced in fusion material, in plasma, in solid , liquid and gaseous substances, etc. In any one of these states there are some variations that here will be omitted, in order not complicate the explanations; only will study the conduction of heat in a solid body (in a flat plate), with all its binary systems not ionized. The heat is transmitted from an atom to the adjacent one because there is a direct contact between them. In order the flow of heat will not be produced in an anarchic way, will consider it is flowing along multimolecules, that at the same time that are distributed parallel to the sides of the plate, covering an area (spiral volume) are in colineal contact with the adjacent spiral volume; also with the same thick of Lm In other themes has been explained that a multimolecule is formed by a chain of 2.5 x 10⁶ atoms; so a transversal section of conduction, a multimolecule could be represented by one atom. In the cross section were are the multimolecule, in which is flowing the heat, transmitted by the electrons. Normal to the trajectory of the flowing particles mh, there is a surface with: L_m² atoms; cross section of spiral volume. (L_m = length of a multimolecule expressed in atoms = 2.5 x 10⁶; each atom correspond to the perpendicular section of a multimolecule. In our model we consider that the spiral surface is formed by: A = 2 x 7 x 8 = 112 multimolecules (see Fig. 1).

Since the beginning of the theme was think that heat in form of particle of corpuscles mh moves along multimolecules. In a flat and extensive plate of uniform thick , in a side of it there is higher temperature than in the other one. The heat will move from the higher temperature side toward the lower temperature, along the multimolecules there are in every spiral volume (see Fig. 1), that are parallel to the sides of the plate. Adjacent with the hotter side of the plate, there is the first spiral volume. At a perpendicular distance L_m = 2.5 x 10⁶ atoms there is other spiral volume (the second one); both are united in the point (0) by a multimolecule that is perpendicular to them, with the values given before. At other similar distance, and collinear, toward the less hotter side of the plate there is a third spiral area. And so on till the less hotter side of the plate.

Considering one spiral volume, in this the heat arrives to point (0) in volume V₁ = A₁ L_m; next it moves a distance L_m to point (1); from here it moves a distance L_m to point (2); next it moves a distance 2 L_m to point (3); from here to point (4) moves 2 L_m; from (4) to (5) moves 3 L_m… from (13) to (14) moves 7 L_m. Then it enters to the pointed line spiral volume A₂, we have moving from (14) to (13) a distance 6 L_m; from (13) to (12) 7 L_m; … from (2) to (1) and from (1) to /0) moves     l Lm; completing a time of spiral. When the fore time of spiral has finished,. from point (0) of its spiral area, the heat moves perpendicular to point (0) of the forward spiral volume at a distance Lm. And so on, till arrives to the last spiral volume adjacent to the hotter side of the plate. To this movement between all the spiral volumes we will call: time of cycle = Σ times of spirals.

With the fore information can obtain some numerical data that will help to structure our model. If the spiral volumes are spaced a perpendicular distance Lm, their influence volume will be a prism with a base equal to A and a height equal to Lm; to this volume will call: spiral volume; in the center of it ought to be a spiral area. We have seen that all the vertex point are marked with a number: 0, 1, 2, …. 14; the fore ones and those indicated with only one point have an equal influence volume of Lm³ atoms. We can imagine that the particles of heat mh can move during a time of jump (7 shell) into the forward electron that transmit the heat and this is yielded to the forward electron, and so on. Also could be consider other model (?) in which the heat advance from electron to electron at a slower velocity of several times of jump / electron, as does the sound.— To the spiral volume that is adjacent to the hotter side of the plate we will call: first spiral volume; in this there are 112 Lm from point (0) of A1, to point (0) of area A2.; of one spiral volume In our model will consider that in a time: 112 t_m = 112 x 2.8 x 19^—5 = 3.136 x 10^—3 Sec. = time of spiral, al the atoms of the first spiral volume absorb 100 % of the heat there is in front of the hotter area; in other time of spiral the fore heat moves to point (0) of the first spiral volume; and in a time: t_m = 2.8 x 10^—5 Sec. = time of spiral jump = tjs, the heat absorbed by the first spiral area has penetrated to the second spiral volume

In the next table, in the first row are indicated with the number 1 to 8 the number of the spira volume that has the metallic plate. In the second row is indicated the quantity of heat that the calorific particle mh yields in each speral in each speral volume, parting from one mh in the first colum: In the third row is indicated the quantity of heaat that the second calorific particle yields in the 8 spiral volume: And so on, in the other rows. Here is considered that when the first calorific particle mh begins to move in the second spiral volume, the second calorific particle mh begins to move in the first spiral volume; and when this begins to move in the second mh the third mh begins to move in the first spiral volume; and so on.

1                 ½                   ¼                        1/8                 1/16                  1/32                    1/64                     1/128
1                 ¼                   ¾                       5/16                3/16                  7/64                    4/64                     9/256
1               7/8               11/16                    8/16               11/32               29/128                37/256                  23/256
 1             5/1                13/16                   21/32               16/32              9 3/216               65/256                  44/256
1            31/32              57/64                  99/128             163/256           128/256             193/512               281/1024
1             63/64             60/64                  219/256            191/256           319/512              256/512              793/2048
1           127/128         247/256                233/256            212/256           743/1024           1255/2048        1024/2048

or to make a numerical example, it will be consider a copper plate of thick: =                                    9 L_m = 9 x 2.5 x 10⁶ (2 x 1.13763 x 10^—10) = 5.1193 x 10^—3 m = 0.51193 cm;                                    L_m = 2.5 x 10⁶ (2 x 1.13763 x 10^—10) = 5.688 x 10^—4 m                                                                 Spiral area: A = A₁ + A₂ = (14 L_m) (8 L_m) = 112 L_m² = 112 (5.688 x 10^—4)²  = 3.623 x 10^—5 m²     Spiral volume = A L_m = 3.623 x 10^—5 x 5.688 x 10^—4 = = 2.061 x 10^—8   m³ = 0.0206 cm³

In the next time of jump of spire: tjs = tm the remaining particle of mh will be in point (0) of the third spiral volume; here will yield half of the remaining mass m / 4, also reducing to the fore value its mass; this process in a time: t_s = 112 t_m: In a time of jump of spire tjs = tm the remaining mass will be in point (0) of the four spiral volume here will reduce and yield to: mh /8. And so on till the 8 spiral volume (adjacent to the less hotter side) in which the reduced and yielded particles will be: mh / 128; completing a cycle.

It has been explained the yield of heat in the sample (the plate) in the first cycle; but after the first heat particle mh was produced in the first spiral volume and had advanced to point (0) of the second spiral volume, all these in a time: 2 ( ts + tjs) = 226 t_m = 113 t_m of absorbing heat + 113 t_m in yielding heat to the point (0) of the second spiral volume; the first spiral volume produces a second mh. So when the first mh is in point (0) of the third spiral volume; the second mh is in point (0) of the second spiral volume: That is, the second heat particle mh will be behind the first one a time of spiral (t_s). -In the second row of the table are given the quantity of unitary heat that get all the spiral volumes in the first cycle, produced by the first heat particle mh . The second heat particle mh will yield to the second spiral volume some heat in a second cycle, as follow:

Second cycle: In the first spiral volume is produced a second heat particle mh proportional to (1), in the third row of the table In a time: (112 + 1) tm, the particle will be in point (0) of the second spiral volume, and will remain during a time ts in this spiral volume; etc… In the second spiral area there is: (1 / 2) mh from the first cycle, plus 1 mh of the second cycle: 0.5 (0.5 + 1) = 3 / 4 = to the deposited heat in the second spiral volume = remaining energy when the first cycle has finished.

Remaining energy of the second spiral volume + Deposited energy of the first cycle; all divided by 2. [1+ 1 /2] 0.5 = 3 /4; [3/4+ 1 /4] 0.5 = 2/4; [2/4 + 1/8] 0.5 = 5/16; [5/16 + 1/16] 0.5 = 3/16; [3/16 + 1/32] 0.5 = 7/64; [7/64 + 1/64] 0.5 = 4/64; [4/64 + 1/128] 0.5 = 9/256

In a third cycle (4 spiral volume) we have:                                                                                      1 + 3 /4] 0.5 = 7/8; [7/8 + 2/4] 0.5 = 11/16; [11/16 + 5/16] 0.5 = 8/16 . . .

In the second column the deposited heat varies: (2 – 1)/2 =1/2;     (4 – 1)/4 =3 /4;      (8 – 1)/8 = 7/8; (16 – 1)/16 = 15/16;    (32 – 1)/32 =31/32;    (128 – 1)/128 0.5 = 127/128.

In the seven cycle (8 spiral volume) we have: 1; [1 + 63/64] 0.5 = 127/128; [127/128 + 60/64] 0.5 = 247/256; . . . . . . . [1245/2048 + 793/2048] 0.5 = 1024/2048

For us that are supporter of mechanophysics, the fore mentioned model resulted completely unsatisfactory; so was obtained the model of the binary systems. As have been explained in other themes of this protocol, with the model of the binary systems the nucleus of the atom is not formed by a unitary particle, but as many particles, as orbital electrons has the atom: This means, that each binary system is formed by an orbital electron as negative particle, and a positive particle formed by a proton; or by a proton and one, or two neutrons. The negative (electron) and the positive particle that form the binary system are spaced one from each other a distance approximate to a radius of an atomic shell. Both particles rotate around a point (center of the nucleus). The orbital electron of the seven shell has, in accordance with our model, a normal velocity of:   v₇ = 308,570 m./ Sec; and the positive particle (proton)has a velocity: v₊₇ = 4 x 308,570 / 1,836 = 672.266 m./ Sec.. If the two mentioned particles of a binary system do not unite, in spite they have opposite electric charges is because there is an equilibrium between the kinetic energy (due to inherent energy…) that has the orbital electron trying to move away in a perpendicular direction from the positive particle; and the attraction force that exist between both particles of the binary system. Similar thing could be say of the positive particle with respect to the orbital electron. To explain with all details all these effects is a long way. Here only will give some direct ideas. In other themes seen in this protocol have been see the binary systems, so, we have some data from they.

An orbital electron moving in the seven shell has a kinetic energy equal to:                                         Ke = 0.5 m_e v₇² = 0.5 x 9.1091 x 10^—31 (3.0857 x 10⁵)² = 4.3366 x 10^—20 joule                                  Mass of a corpuscle:     m_c = 1.47236 x 10^—50 Kg.                                                                            We can consider the fore energy equal to the inherent energy of the electron. With this energy the electron moves away from the positive particle in a perpendicular direction. The inherent or kinetic energy of the electron could be produced by (n) active corpuscles, as those of light, moving at light velocity, so:           Kinetic energy of a corpuscle in a propeller fluid:                                                 K_e  =  0.5 m_c c² = 0.5 x 1.47236 x 10^—50 (3 x 10⁸)²  = 6.62562 x 10^—34   joule  =  one quantum

In many other examples have been obtained the deflection velocity (radial velocity) of the negative particle (electron). v_^7 = 69,098 m./ Sec..; the deflection angle:   α₇  =  36^0o /28 = 12^o.857                 K_^+7  =  0.5 x 9.1091 x 10^—31 x 69,098² = 2.1746 x 10^—21 joule.

The orbital velocity of the proton: v₊₇ = 672.21 m./ Sec.                                                                  Its radial velocity: v_^+  =  69,098 x 672.21 / 308,570 = 150.54 m./ Sec.                                             K_^+  = 0.5 m₊ v_^+² = 0.5 x 1.6725 x 10^—27 x 150.54² = 1.896 x 10^—23 joule                                        K_^+7 / K₊ = 2.1746 x 10^—21 / 1.896 x 19^—23 = 114.7

If the (m_p) acted in the same way in the positive particle than in the negative one: why the attraction forces K_^+7 and K_^+ are so different? With the following law will be explained why.                         1.896 x 10^—23 x 1.836 / 4² = 2.1756 x 10^--21 = K_^+7

Law of interchange of an inherent energy: The interchange of inherent energy is 4 times more effective or intense in the positive charges than in the negative ones, and is proportional to the quantity of inherent energy that has the propeller particle (mp) divided by the active inherent energy that has the affected particle (in this case the proton).

Monterrey, Mécico, April, 2004 Manuel de Hoyos Robles.

In the binary systems, although the electron is the smaller particle, is the most active one; in a more direct way, with it will be define the behavior of such system and of the atom; so, by preference will study its structure and behavior. Beside the inert corpuscles that form its structure with a mass: m_e = 9.1091 x 10^—31 Kg, the stable electron has, in accordance with our model some free corpuscles whose inherent energy is equal to: 13.5 eV., part of these corpuscles are employed in produce the orbital velocity of the electron (kinetic energy: 0.5 me vn²); the other part is stored into the electron of the binary system) we call this energy: stored energy

In our models of molecules and multimolecules an electron jumps from an atom to the adjacent one every time of jump:   t_j = 1.12 x 10^—11 Sec. With this we conclude that for to produce an ionization (in the seven orbit) it is required the fore time; this is one condition. After was made the model of electric current was consider that the current electron could jumps from a multimolecule to the adjacent one, producing a big ionization; but when all the values of the electric current were balanced, the required ionization was reduced in a signify way Afterward we study better the model. was seen that the fore ionization only could be produced if the current electron moves several times of jump (tj) around each atom of the multimolecule. For to solve this problem we have two ways: One considering that the current electron moves several times of jump only in the last atom of the multimolecule, before it jumps to the adjacent multimolecule. The other solution is considering that the atoms of the multimolecules get more corpuscles than the corresponding to the stored energy of an isolate atom, so they needed less energy to get ionized.. This solution is the correct one because does not affect the times in which are produced the different effects… Forward will be seen that the electron of binary system (seven shell) can gets more independent corpuscles than the stored ones; for instance, when it is affected by an ionization medium. These corpuscles can be obtained because the other binary systems of the atom yield them to such binary system. Also it is possible that these extra corpuscles could be formed at expense of the same structure of the electron of the binary system, when this is affected by the ionization effect. By now it does not matter how they are formed, only matter that they exist This extra corpuscles can produce an extra energy in the electron that we will call: deposited energy. and that is manifested when the ionization is produced. All these could be understand better in the numerical examples.

The energy of ionization is reduced in a perceptible way in the electric current electrons, as was seen in the fore paragraph see the theme: A Model of Electric Current and its Application to Medicine. If the atoms of an electric conductor have more deposited energy (deposited corpuscles) than the normal one (or stored energy), without growing their temperature, this poses some interesting problems in physics and in biology. As the binary system is formed by a positive particle and a negative one (electron), this last one, when acts in the molecules and in the multimolecules, jumps from a binary system of an atom to the adjacent atom, every time o jump.

A stable atom can gets unstable if is affected by an ionization medium. Because this can imagine that the ionization medium emits corpuscles that are absorbed by the electron of the binary system is going to be ionized. We rather think that the ionizer medium emits a field of microcorpuscles that affect the electron, in such way that the other binary systems of the atom, corresponding to shells smaller than the seven one, contribute to increment the corpuscles forming the deposited energy; this could be done with corpuscles emitted by the other binary systems. In our model we will see here, the only thing that interest us, by now, is to know the quantity of deposited energy; and the potential of ionization; this last one will be obtained with the values given by the tables. Forward will be given a formula, with a mathematical criterion for to determine the structure of the atoms, based in their ionization potential and of the radius of the atoms.

When an atom is affected by a difference of potential, can get ionized, this happens in one of the two binary systems of the seven shell. In our models we will see here, also there are some condition to produce such ionization; one of them is that the axis of the binary system is going to be ionized, be oriented in the same direction in which is the ionizer medium with respect to the atom; that is, that the electron will be in front (see propen side forward) of the ionizer medium in the instant that the ionization begins. This ionization effect continues acting in the electron in every vertex of its orbits during a time of jump, and then the electron jumps out from the binary system that is ionized.

About 40 years ago I conceived the model of the polygonal orbit atoms, with this I was forced to consider the velocity of the orbital electrons; in the first orbit the velocity was equal to 2,160 Kms./ Sec. With this velocity the electron got a kinetic energy equivalent to 13.3 eV.; that was the energy required for the orbital electron to jump from the seven orbit (shell) to the first one. In the seven orbit the kinetic energy of the orbital electron was consider: v₇ = 2,160 / 7 = 308.57 Kms./ Sec., that is equivalent to an energy equal to: K₇ = 13.3 / 7² = 0.27 ≈ 0.3 eV. In our model of polygonal orbit were forced to consider the seven orbit (7 shell) with 28 sides. Then (not now) it was consider that the time of vertex could vary in the same proportion than the fundamental radius did; since then was given to the radius of the atoms an average a value:   r _F = 1.13763 x 10.¹⁰ m.

Here will be consider that all the electrons dispose of an energy equivalent to 13.6 eV.: A seven orbit electron employs 0.27 ≈ 0.3 eV. of the fore energy for to move at: 308.57 Kms./ Sec, and this and the rest of such energy is stored one. From the six to the second shell the binary systems have the same stored energy, so can contribute (in accordance with our model) with part of these energy to increment the deposited energy of the binary systems that are going to be ionized (seven orbit ones), in a complete way the first binary system and in less magnitude to the second binary systems. In the first binary systems all the stored energy of it is employed in produce its velocity: v₁ = 2160 Kms./ Sec. This binary system does not contribute to increment the deposited energy of the binary system that is going to be ionized. Here it is pertinent to point out that all atoms (except the H one) have two orbital electrons in the seven shell; one that can be ionized (first binary system) and the other (second binary system) that can receive some deposited energy (±) in this process, although in a smaller quantity.

Considering that the times of vertex are a constant value in all the atoms, as was proved in the theme: Unified Fields, we will obtain a simplified formula of ionization; with this formula, not only will be able to explain how are formed the atoms, but also will be able to understand many properties of them. The numerical results will be better explanations that all the words we use. When the radius of an atom grows, the velocity of the seven orbit electron will grows in the same proportion and the kinetic energy of such electron will grows in a proportion equal to the square of such velocity. The increments or decrements will be given considering a unitary value with respect to the fundamental radius  r_F. As was say before, in an electron there is a stored energy equal to 13.6 eV. A part of the fore energy is employed in giving the orbital velocity to the electron. Considering the fundamental radius, the normal or fundamental kinetic energy of the orbital electron is as follow: Seven shell: K₇ = 13.6 / 72 = 0.278 eV. Sixth shell: K₆ = 13.6 / 6² = 0.378 eV.   K₅ = 13.6 / 5² = 0.544 eV.   K₄ = 13.6 / 4² = 0.85 eV.   K₃ = 13.6 / 3² = 1.511 eV.   K₂ = 13.6 / 2² =3.4 eV:    In the first shell all its stored energy is employed in moving the electron in such orbit.

We have say that any electron has a total stored energy of 13.6 eV; of this energy, a binary system corresponding to the seven shell employs K7 energy in moving the electron in the seven orbit at a velocity v₇, and the rest of the energy: 13.6 - K₇ = K_7s = stored energy that remains in the electron. Beside the fore energy the seven orbit binary system has other energy that contribute to produces an ionization the electron; this is the deposited energy (energy given by the other binary systems) plus the ionization one, both are equivalent to 13.6 eV. It is supposed that the deposited energy is produced by some binary systems, as have been explained; nevertheless, yet we are not sure if the stored energy contribute in some way to this. The energy that receive the first and the second binary systems when are ionized is called: deposited energy that yield the other binary systems of the atom, when is ready to be ionized, they are: K₇  =  13.6 / 7²  =  0.278 eV.  Sixth shell:  K₆   =  13.6 / 6²  =  0.378  eV.  K₅  =  13.6 / 5²  =  0.544  eV.;   K₄  =  13.6 / 4²  =  0.85  eV.   K₃  =  13.6 / 3²  =  1.511  eV.   K₂  =  13.6 / 2²  =3.4  eV             eV.;     K_2d = 13.6 -3.4 = 10.2

Δrn .In order not complicate too much our explanations, in all these will not consider the term Δ rn; also in obtaining the formula of ionization, for the same reason. In the numerical examples obtained with the simplified formula, we will be able to correct the approximate results. Of the two orbital electrons of the seven shell one will be ionized due to an exterior incitation that in the annex table is called: ionization potential (= i), that is the energy (plus the deposited energy…) that by itself the binary system needs for to be ionized. But the adjacent binary systems of the atom can give to the binary system that is going to be ionized some energy that we will call: deposited potential = K_d, so that: K_i + K_d = 13.6 eV

In the table of ionization can be appreciated that in the extreme right column the ionization is bigger than 13.6 eV .in the atoms: He, Ne, Ar, and Kr. In the rows in which are the fore mentioned last three atoms, the value of ionization grows in an average way from the extreme left atom toward the extreme right atom. When the ionization of the atoms is bigger than 13.6 eV, one binary system can get ionized and the other also increments its deposited energy, but in a smaller quantity. In a row, of the fore mentioned, from the left atom toward the extreme right atom, in an average way the ionization grows as will be seen forward.

In the (Xe) atom the ionization is: i = 12.1 eV., in the (Rn) atom is: i = 10.7 eV. This     means that the second binary system decrements its stored energy when the first binary system is reaching its ionization energy; but this energy, by one side grows in accordance with the ionizer  medium, and by other side diminish in accordance with the capacity of the atom to gets deposited  energy. This means that as the amu of the atom grows, its capacity to gets deposited energy  diminish; in this way can be explained that could not be atoms with higher amu = 102.    As have been explained, the binary system that is going to be ionized of an atom, gets deposited  energy of all the other binary system of the atom, in a proportion equal to their stored energy they  have, a binary systems of the first shell does not have free stored energy because it is employed all  this energy in moving the electron at a velocity: v₁ = 2,160 Kms./ Sec.

Next will be given the proportion of stored potential obtained by each binary system of each complete shell: Seven shell, in this the binary system (second one), that is partially ionized contributes with a proportion equal to: (7 / 7)². In the sixth shell, each binary system contribute with a proportion equal to: (6 / 7)². All the binary systems corresponding to the sixth shell contribute with a proportion equal to:   (6 / 7)² x 8. In the fifth shell all the binary systems corresponding to this shell contribute with a proportion equal to: (5 / 7)² x 8. In the third shell = 5 row: (3 / 7)² x 18 In the second shell (6 row) the proportional contribution is: (2 / 7)² x 32 (there are 32 atoms in the 6 row). It has been say that all the binary systems (minus , that,t hat is going to be ionized) contribute to produce the deposited potential. This potential will be equal to the sum of all the binary systems that contribute divided by 13.6 (N – 1), and multiplied by (K_d)²

The atoms have 7 shell, but not all are occupied by binary system in all the atoms; such shells are denominated by the numbers: 1, 2, 3, 4, 5. 6. 7. But not all the orbital electrons move exactly in the whole number shell mentioned before: Most of them move in orbits (subshells) with radiuses little bed bigger (s, p, levels), or little bed smaller (f, d, levels) (?)

Next will be determine the simplified formula of ionization, with all say before                             13.6 + i²´ = i + A·B·C . . . . (I)

In the ionization of an atom, all the binary systems work in some way. In our model, the two binary systems of the seven shell are those that are affected in a direct way by the ionizer medium; one that we will call: first binary system will gets from the other binary systems the energy required for gets ionized; and the other we will call: second binary system will gets a fraction of ionization. The first binary system (7 shell) gets an energy of 13.6 eV. (equal to the ionizer energy plus the deposited one) The second binary system will take the other fraction of the ionization, that is represented by: i²´

(A), this term is is equal to: (r_N / r_F)²;      r_N = radius of the atom;      r_F = fundamental radius = 1.13763 μ.. Have been explained in other part of these theme that the energy that has the seven orbit electron grows proportional to the second power of the increment of its radius.

(B)

this term is equal to: K_Nd² / 13.6 /N – 1). The value: K_Nd / 13.6 is a proportion of the deposited potential that has the smaller binary syst4em of the atom. So, the fore proportion is the maximum that can have such system. And as we consider that all the binary systems contribute with an equal proportion each one, the fore proportion will not vary in the other binary systems; the only thing will vary is the capacity to yield energy to the first and second binary systems, as will be seen forward. K_Nd / 13.6 is the average quantity of energy that receive the first and second binary systems (seven shell) from the other ones; in a mks system the fore value is expressed in joules.

(C)

the terms into the angular parenthesis: (n/7)² are proportional to the energy that each binary system contribute to produce the deposited potential of the first and the second binary systems. The value 8´ represents the quantity of binary systems there are in the six shell. The value 8” represent those that are in the 5 shell. The value 18´those that are in the 4 shell. The value 18” those that are in the 3 shell. The value 32´ those that are in the 2 shell. We have: 8´+ 8” + 18´ + 18” + 32´ = N – 1;   being N equal to the number of binary systems that has the atom. The average contribution of each binary system will be: C / (N - 1). The value (N – 1) is a denominator one in the term (B). In the formula (I) of ionization, A·B·C gives the deposited energy; and (i) gives the potential of ionization.

In this theme has been talked of a fundamental radius of the atoms: rF; with this radius the orbital electron (7 shell) moves at a velocity: v₇ = 2160 / 7 = 308.57 Kms./ Sec.: If the atom has a radius different than the fundamental, the velocity of the electron and its kinetic energy varies has have been explained

(Be ) atom.    i = 9.3 eV;      r_N = 0.89 μ;      N = 4 ;     M = 6 shell;       K_6d = 13.2 eV.                  A =  (r_N / r_F)²  =  (0.89 / 1.13763 )²  =  0.782²  = 0.612                                                                   B =  K_6d² / 13.6 (N – 1)  =  13,2² / 13.6 (4 - 1) = 4.271                                                                  8´ = 2; 8” = 18´ = 18” = 32´ = 0                                                                                                   C = [ (7 / 7)² + (6 / 7)² x 2 + 0 ] = 1 + 1.469 = 2.469                                                               Formula (I): i₁         1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                          13.6 + i²´ = 9.3 + 0.612 x 4.271 x 2.469 = 9.3 + 6.454 = 15.754
The value:    i²´ = 15.754 – 13.6 = 2.15 = degree of ionization of the second binary system.
A·B·C = 6.454 eV. = deposited energy that receives the first and second binary systems.

(C) atom.    i = 11.3 eV;      r_N = 0.77 μ;      N = 6 ;       M = 6 shell;       K_6d = 13.2 eV.
A  =  (r_N / r_F)² = (0.77 / 1,13763)² = 0.577² = 0.458
B  =  K_6d² / 13.6 (N – 1) = 13.2² / 13.6 (6 - 1) = 2.562 eV
8´ = 4; 8” = 18´ = 18” = 32´ = 0
C = [ (7 / 7)² + (6 / 7)² 4 + 0´ ] = 3.939
Formula (I):  i´    1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C

(N) atom.    i = 14.5 eV;      r_N = 0.74 μ;     N = 7 ;      M = 6 shell;      K_6d = 13.2 eV.                      A = (r_N / r_F)² = (0.74 / 1.13763)² = 0.650² = 0.423                                                                          B   =  K_Md² / 13.6 (N – 1) = 13.2² / 13.6 (7 - 1) = 2.135                                                                  8´ = 5 8” = 18´ = 18” = 32´ =  0                                                                                                  C = [ (7 / 7)² + (6 / 7)² 5 + 0 ] = 4.673                                                                                     Formula (I): i´    1 + i´² = i + A·B·C  =  13.6 + i´² = i + A·B·C                                                         13.6 + i²´       1 =  4.5 + 0.423 x 2.135 x 4.673 = 14.5 + 4.220 = 18.720

(Na) atom.    i = 5.1 eV;      r_N = 1.57 μ;      N = 11 ;      M = 5 shell;     K_5d = 13.1 eV.                     A = (r_N / r_F)² = (1.57 / 1.13763)² = 1.380² = 1.905                                                                       B = K_5d² / 13.6 (N – 1)  =  13.1² / 13.6 (11 - 1) = 1.262                                                                 8´ = 8; 8” = 1; 18´ = 18” = 32´ = 0                                                                                               C = [ (7 / 7)² + (6 / 7)² 8´ + (5 / 7)² 1 + 0] = 7.388                                                                     Formula (I): i´       1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                        13.6 + i²´       5.1 + 1.905 x 1.262 x 7.388 = 5.1 + 17.79 = 22.89

(K) atom.    i = 4.3 eV;      r_N = 2.03 μ;      N = 19 ;      M = 4 shell;      K_4d = 12.75 eV.                    A = (r_N / r_F)² = (2.03 / 1.13763)² = 1784² = 3.184                                                                          B = K_4d² / 13.6 (N – 1) = 12.75² / 13.6 ( 19 - 1) = 0.664                                                               8´ = 8” = 8; 18´ = 1; 18” = 32´ = 0                                                                                              C = [ (7 / 7)² + (6 / 7)² 8 + (5 / 7)² 8 + (4 / 7)² 1 + 0´] = 11.287                                                 Formula (I): i´   1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                            13.6 + i²´     1 + 4.3 + 3.184 x 0.664 x 11.287 = 4.3 + 23.860 = 28.161

Here: i²´ = 28.161 – 13.6 = 14.56 eV. We expected i² would be smaller than 13.6 eV. The interpretation we give to this is that when is produced the ionization effect, the second binary system of the seven shell, moves to a subshell smaller than the seven one; so its potential of ionization grows; forward, in the atom of copper will be seen with more detail this.

(Cu) atom.     i = 7.7 eV;      r^N = 1.17 μ;      N = 29 ;      M = 4 shell;      K_4d = 12.75 eV.                A = (r_N / r_F)² = (1.17 / 1.13763)² = 1.028² = 1.058                                                                           B = K_4d² / 13.6 (N – 1) = 12.75 ² / 13.6 (29 - 1) = 0.427 8´                                                             8” = 8; 18´ = 11; 18” = 32´ = 0                                                                                                       C = [ (7 / 7)² + (6 / 7)² 8 + (5 / 7)² 8 + (4 / 7)² 11+ 0] = 14.552                                                 Formula (I): i´     1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                         13.6 + i²´ = 7.7 + 1.058 x 0.427 x 14.552 = 7.7 + 6.574 =14.274

We can find a very much important fact if we see the theme: A Model of Electric Current and its Application to Medicine, in which is given a numerical example of electrical conductor of copper wire. A copper atom gets ionized when it is applied a potential of 7.7 eV.; because this, can suppose that for produce an electric current it is required to apply the fore mentioned difference of potential; nevertheless, in such example with a difference of potential equal to: ik = 0.192 eV is produced an electric current. The interpretation given to this is that the second binary system of the seven shell, contribute whit an energy of: 7.7 – 0.192 = 7.508 eV. (absorbed by the first binary system) when is produced the difference of potential required for produce the electric current. So, will have that for an electric current will vary some values: i²´ = 13.6 + 7.508 = 21.11 eV; i = 0.192 eV; formula (1) will be transformed: 13.6 + i²´ = 0.192 + A B C

(Rb) atom.     i = 4.2 eV;      r_N = 2.16 μ;      N = 37 ;      M = 3 shell;      K_3d = 12.1 eV.                  A = (r_N / r_F)² = (2.16 / 1.13763)² = 1.899² = 3.605                                                                          B = K_3d² / 13.6 (N – 1) = 12.1 ² / 13.6 (37 - 1) = 0.299                                                                   8´ = 8” = 8; 18´ = 18; 18” = 1; 32´ = 0                                                                                        C = [ (7 / 7)² + (6 / 7)² 8 + (5 / 7)² 8 + (4 / 7)² 18 + (3 / 7)² 1 + 0] = 17.022                                 Formula (I): i´     1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                         13.6 + i²´ = 4.2 + 3.605 x 0.299 x 17.022 = 4.2 + 18.348 = 22.548;        i²´ = 8.95

(Sr) atom. i = 5.7 eV; rN = 1.91 μ; N = 38 ; M = 3 shell; K3d = 12.1 eV.                                         A = (rN / rF)² = (1.91 / 1.13763)² = 1.679² = 2.819                                                                        B = K_3d² / 13.6 (N – 1) = 12.1² / 13.6 (38 - 1) = 0.291                                                                   8´ = 8” = 8; 18´ = 18; 18” = 2;   32´ = 0                                                                                       C = [ (7 / 7)² + (6 / 7)² 8+ (5 / 7)² 8 + (4 / 7)² 1´ + (3 / 7)² 2 + 0] = 17.205                                 ¨Formula (I): i´     1 + i´₂ = i + A·B·C = 13.6 + i´² = i + A·B·C 13.6 + i2´ = 5.7 + 2.819 x 0.291 x 17.205 = 5.7 + 14.114 = 19.814

Cs) atom.     i = 3.9 eV;      r_N = 2.35 μ;      N = 55 ;      M = 2 shell;      K_2d = 10.2 eV.                    A = (r_N / r_F)² = ( 2.35 / 1.13763)² = 2.066² = 4.267                                                                        B = K_Md² / 13.6 (N – 1) = 10.2 ² / 13.6 (55 - 1) = 0.142                                                                  8´ = 8; = 8; 18´ = 18” = 18;   32´ = 0                                                                                            C = [ (7 / 7)² + (6 / 7)² 8 + (5 / 7)² 8 + (4 / 7)² 18 + (3 / 7)² 18 + (2 / 7)² 1] = 20.226                 Formula (I): i´     1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                         13.6 + i²´ = 3.9 + 4.267 x 0.142 x 20.226 = 3.9 +12.255 = 16.155
i²´ = 16.155 - 13.6 = 2.555

(Hg) atom.     i = 10.4 eV;      r_N = 1.44 μ;      N = 80 ;      M = 2 shell;      K_2d = 10.2 eV                  A = (r_N / r_F)² = (1.44 / 1.13763)² = 1.266² = 1.602                                                                         B = K_2d² / 13.6 (N – 1) = 10.2² / 13.6 (80 - 1) = 0.0968                                                                  8´ = 8” = 8;; 18´ = 18” = 18;   32´ = 26                                                                                         C = [ (7 / 7⁾² + (6 / 7)² 8 + (5 / 7)² 8 + (4 / 7)² 18 + (3 / 7)² 18 + (2 / 7)² 26] = 22.266                 Formula (I): i´     1 + i´² = i + A·B·C = 13.6 + i´² = i + A·B·C                                                         13.6 + i²´ = 10.4 + 1.602 x 0.0968 x 22.266 = 10.4 + 3.453 = 13.73 ≈ 13.853

Monterrey, Mexico, February 2005
Manuel de Hoyos Robles.

Un corto de vista necesita sus lentes para buscar sus lentes y poder ver bien, Esto lo digo porque el Dr. José Manuel Ramírez Aranda y yo tenemos una idea de cómo erradicar la enfermedad del VIH, y es muy fácil comprobarlo sin ningún riesgo, con la electroterapia y la ayuda oportuna de pruebas de laboratorio. La mas reciente terapia de los tres medicamentos simultáneos, lo único que está haciendo a largo plazo , es producir virus mas resistentes.

In the theme: How are Formed the Atoms, were seen some important properties that they have. It was defined there, some important and fixed values, as the time of vertex: tv7 = 8.27 x 10^—17 Sec.; the stored energy that has each orbital electron and that is equal to: 13.6 eV.; the fundamental radius of the seven shell equal to: r_F = 1.13763 x 10^—10 m., As all the electrons are similar, all them ought to behave in an equal form when they are affected by equal effects In all atoms, that are formed by binary systems, the positive and the negative particles always try to behave in similar way. The orbital electrons try to move around the whole number shells (1, 2, 3, 4, 5, 6, 7 shells), with a velocity: vn = v1 / n = 2,160 / n Kms./ Sec.: The attraction (in the binary systems) between the positive particles and the negative ones is more effective than the rejection forces between the particles with the same sign. All these is produced without affecting the invariant values mentioned at the beginning of this theme. If by any reason can not be realized the fixed values of the atoms they would be disintegrate. This means that all the effects produced in the atoms have their limitations in space, time, energy, etc.; but also can be some variations that in some way is as a liberty for the atomic particles; in this way an orbital electron can move in a subshell, in accordance with the condition of the atoms, their temperature, the chemical reaction, and other facts, as will be seen here

In our design of the model of molecule will attach to all have been say before. Maybe some times will not understand what produces some effects, but we will be very much strict in accept it if fulfills the required condition. By now we will limit to study the most elementary molecules formed by two or three elements. If we are strict in accepting the fixed and invariant values, in a molecule formed by two elements (two atoms), both atoms ought to have their seven shells with equal dimension of their radiuses.

In the theme: How are Formed the Atoms have been worked with the different radiuses that have the atoms, but it was seen that there are some variations in such radiuses due to the ionization potential, to the deposited potential, that here will denominate: balanced potential, because as will be seen in the numerical examples, in the same way that helps to the ionization potential to take out an electron of an atom; could help to the ionization potential to introduce the electron to other atom (to the other atom of the molecule). This could be interpreted as between the balanced potential and the ionization potential can be a change or variation of their properties; in some conditions the balanced potential could be transformed in ionization potential, and vice versa, the ionization potential could be transformed in balance potential.

In the table of radiuses are given the dimension of them, when they are in form of elements in a stable state. In an isolated atom there is an interchange of energy between all the parts that form it. In a molecule also there is some interaction between all the atoms that form it. Here will give a model of molecule formed by two or three atoms. For to form the molecule, being respectfully of the rules have been determine in the explanation of other phenomena; then the atoms are forced to have the same radius. Also will be consider there will be some interchange between the ionization and the balanced potentials.

In a molecule formed with two atoms, in a given instant, that is equal to the time of jump (7 shell): one of the atoms is as an ionized one; that is, a binary system of it has not its negative particle, or orbital electron, or binding electron that is moving in the other atom. In the next time of jump the binding electron will be moving in the first mentioned atom; and the second atom will be as an ionized one, and so on, happens in each time of jump. Of course this is verify in one of the two binary systems corresponding to the seven shell in each one of the two atoms Originally it was consider that the molecable or binding electron, jumped from one atom to the other one of the molecule in a time of orbit, describing a trajectory similar to an eight (8). When the electron moves in one atom, its corresponding binary system employs its energy in deflects the trajectory of the electron; in the other atom the positive particle of its active binary system without its negative particle, accumulate energy during a time of jump: tj; in such way that when the binding electron is in a coincident side of both atoms, the electron changes its trajectory from the active atom toward the ionized one; and so on, is repeated this process and change from an ionized atom to an active one, and vice versa. In the coincident side of both atoms, the active one is affected by the ionization potential produced by the ionized atom; and the ionized atom, with the balanced energy accumulated. during a time of jump attracts the binding electron to its orbit. This means that a alanced potential can be transformed in an ionization potential when the balanced ones tends to grows and the ionization one tends to diminishes; and vice versa, an ionization potential can be transformed in a balanced one when is produce an opposite effect. It have been say in other paragraphs that the sum of the deposited and the ionization potentials, are equal to: 13.6 eV. In this way in both atoms the deposited energy is different; but the sum of them ought to be equal to 13.6 eV.

The atom that was active, has accumulated during its active time a balanced potential, that added to the ionization potential given by the no active atom, produces in it a saturated potential, equal to 13.6 eV.; After a time of jump this process is reversed This means that the balanced potential of the second atom, plus the ionization potential of the first atom given to the second one are equal to 13.6 eV.; and vice versa; this in accordance with our model.

The ionization potential of an atom, apparently is mainly a function of the size (radius) of an atom: we have for one atom that its: balanced potential / ionization potential = 1.00. The variation of the size of each atom, mainly is function or the ionization potential of the atom, but we suppose that the ionization also gives other characteristics to the atoms, as the chemical, electric and other properties; so the function r², ionization, only is an approximate one For the atoms to form a molecule the fore condition must verify. The ionization potential of one atom ought to be equal to its balanced potential. The balanced potential of an atom plus the ionization potential of the other atom affecting to the first one is equal to 13.6 eV. equal to a saturation potential. All these explanation will be clear up with the numerical examples; and with this, the chemistry will be transformed from a recipe technique to a scientific one.

In our model of molecule with two atoms it is considered, that both have the same radius.

Molecule of Cl Na;

₁₇Cl³⁵;      r_Cl = 0.99 μ;          i_Cl = 13.0 eV.                                                                             ₁₁Na²³;      r_Na = 1.57 μ;      i_Na = 5.1 eV:                                                                                         First will be determine the molecular radius: r = 0.5 (0.99 + 1.57) = 1.28 μ. 1.28² = 1.6384 Cl:          1 = r² (Z / A)³ i_Cl = 1.28² (Z / A)³ i_Cl = 1.6384 (17 /35)³ i_Cl = 1.6384. x 0.1146i_Cl;                  i_Cl = 1 /0.1878 = 5.326 eV.                                                                                                       Na:     1 = 1.6384 (11 / 23)³ i_Na = 1.6384 x 0.1094 i_Na = 0.1792 i_Na;                                                    i_Na = 1 / 0.1792  =  5.579 eV.

If in modern physics it is consider that light is a universal constant: c = 3 x 10⁸ m / Sec. they are not able to explain why light moves at lower velocity in a refringent body; also they could not explain why some quasars can expand at velocities as big as 9 c. In a simple way can be explain all these, saying that light is not a universal constant. But if in most of the experiments and observations have obtained values of the velocity of light equal to: c = 3 x 10⁸ m / Sec. the fore given affirmation results problematic. In other themes have been say that the corpuscles as those of light move from zero velocity and get the light one with out any force, for reason ignored by all us. Other thing can be say is that if light is emitted from a body that is moving at (v) velocity, grts: c +^> v    After a lapse of time or of distance, that could be a differential one, the light radiation always try to get the (c) velocity; this in accordance with the quasars observations, that expanded at 9c, and were observed at light velocity (c); and with the dynamic of the corpuscles and microcorpuscles. In this theme will be see some reasons why the corpuscles change their velocity, for instance, into a refringent medium.

When a ray of light penetrates into a refringent medium, the superficial atoms affect such radiation, reducing its velocity; not so, the more interior ones, as will be seen in our numerical examples. The corpuscles of the radiation that were affected by the superficial atoms get saturated for some time, we have consider equal to a multimolecular one: tm = 2.8 x 10^—5 Sec. Between other reason, this is in accordance with our model of electric current… In such way that all light radiation that penetrate in the refringent medium remain with the same refringent conditions, in the fore mentioned time. In other themes have say that the corpuscles as those of light, as gravity microcorpuscles have the same energy if they move slower, as light velocity; in the first case their energy is latent, and in the second one is manifested as a kinetic one. With the fore explanation it is easy to understand the behavior of light in a refringent medium. Imagine a recipient full of water, and a ray of light; before it enters to the recipient has a velocity equal to (c ) and its corresponding kinetic energy; when it enters to the recipient gets a velocity: c_r < c, and its corresponding kinetic energy A modernist physicist can not explain this because for him the reduction of velocity produces a lost of energy, and the new increment of velocity when the ray gets out of the refringent medium will not produce an increment of energy. The problem will be understand better with the numerical examples.

Galileo and Newton established the laws of the dynamic of the macroscopic bodies and particles by the observation of them; but they never thought in the origin and roots of the movements. For Einstein the concept of energy was more fundamental than that of movement. For us the differences of movement of the bodies and particles can produce the energy. For him the matter could be transformed in energy: E = mc². Here the term E has no meaning without movement of m…

In the XIX century Tyndall supposed that the sky gets its blue color because in the atmosphere are some dust particles that can disperse some of the rays that move over the atmosphere. He did not explain why the sky gets red when the Sun is near the horizon. Here will be given other explanation, considering why it gets blue when the Sun is up from the horizon, and gets red when it is near the horizon; and why it does not get the other color of the rain bow. Here will work with variation of kinetic energy.

An atom with r = 1.13763 x 10^—10 m, of its seven orbit electrons, has a kinetic energy of each one: K₇ = 0.5 m_e v₇² = 0.5 x 9.1091 x 10^—31 x 308,57 0² = 4.33586 x 10^—20 joule In our model de fore value is expressed in corpuscles moving at light velocity (quantums). K7 / h = 4.33586 x 10^—20 / 6.6256 x  10^—34 = 6.545 x 10¹³ h / Sec.

When all the corpuscles of a radiation pass near an atom or a molecule of a refringent medium they are affected by the fore quantity in some of the atoms or molecules of the refringent medium, as will be explained in the numerical examples.. For determine the refraction of light it is required to have the refraction index; this has been obtained by experiments; in the book: Physics, Principles and Problems, by James T. Murphy, and Robert C. Shoots are given some values of it: water. 1.33; alcohol, 1.36; quartz, 146; polyethylene, 1.52; glass flint, 1.61; glass crown, 1.62; diamond, 2.42. In accordance with modern physics this index is produced by a force; but in the dynamic of the corpuscles and microcorpuscles the concept of force or of energy results an ambiguous one; and now we have to repeat the idea that gives the fundamental of the Universe, the fore elementary particles, without any force or energy can be stable or moving at light velocity. with the same energy. In the atoms, the molecable electrons produce the exterior kinetic energy.

It was seen before that a molecable electron of an atom in its seven shell is affected by a kinetic energy equal to K₇; in the same way could be affected the radiations that pass near it. For to understand better it is necessary to obtain the value of K₇; the seven shell molecable electron can move at v₇ velocity during indefinite time because every time of vertex: t_v7 = 8.27 x 10^..17 Sec. is produced a deflection energy K_^7 = 0.5 me v_^7² = 0.5 x 9.1091 x 10^—31 x 69100² = 2.1747 x 10^—21 joule / t_v7 = 2.1747 x 10^—21 / 8.27 x 10^..17 = 2.6296 x 10^—5 joule / Sec. K_^7 = 2.1747 x 10^—21 / 6.6256 x 10^—34 = 3.2823 x 10¹² h / t_v7

The fore could be interpreted as the atoms have the property or capacity to produce the fore energy every time of vertex (t>7), so, if a radiation as light passes near an tom or into its shells will be affected by such property or capacity. If the corpuscles of the ray of light are affected as was say, after that, they can continue moving into the refringent medium without been affected by the other atoms of such medium, during a time of multimolecule (t_m = 2.8 x 10^—5 Sec,), as was say before. In the fore time the ray moves a big distance with its single refracted velocity, in the atmosphere. A refracted ray of light of the Sun, when moves up in the sky produces a blue color, and when the Sun is near the horizon, produces a red color. All these is because of the big quantity of rays we receive, a small quantity of them are affected by the atoms of the atmosphere. Of all the radiation received from the Sun we can perceive by our eyes those with: λ = 4 x 10^—7 to 7 x 10^—7 m    ν₄ = c / λ₄ = 3 x 10⁸ / 4 x 10^—7 = 7.5 x 10¹⁴ h / Sec.    (violet color) ν₇ = c / λ₇ = 3 x 10⁸ / 7 x 10^—7 = 4.2857 x 10¹⁴ h / Sec. (red color) The average ν is:   ν = (ν₄  ν₇)^0.5 = (7,5 x 10¹⁴ x 4.2857 x 10¹⁴)^0.5 = 5.669 x 10¹⁴ h / Sec.        In accordance with the table of radii of the atoms, the average one is: r = 1.0 μ. It was say before that the kinetic energy that has a molecable electron (7 shell) is produced by the velocity; v₇ = 308,570 m / Sec., and that this velocity is maintained due that each time of vertex is produced a radial

energy equal to K_^7. It was explained before that a refracted ray of light, with some characteristics…will be seen forward, gets a decrement of its kinetic energy equal to K_^7 when all its corpuscles are affected by the fore energy of the atoms of the refringent body. Here it is suppose that the atoms of a solid or a liquid, are in contact one with each other. In a gas, or in the atmosphere the atoms are spaced at relative big distances, beside they are not fixed in a place, but are moving, because this, when a corpuscle of the light radiation gets its corresponding energy of the K_^7, this energy will work in an active atom in the gas, and will be proportional to K₇, instead of K_^7. In an average way the atmosphere is formed by molecules of N2; the radius of N is equal to: r_N = 0.74 μ; considering there are 4 molecable electrons in a N2 molecule. The energy of K₇ vary proportional to (r_μ²) = (r_μ / r)²; here: r = 1 μ.; all the values are expressed in equal magnitude. r_μ = 0.74 μ. The kinetic energy of a molecule 0f N2 is equal to: K₇´ = 4 K7 r_μ² = 4 x 6.545 x 10¹³ (0.74)² = 1.4334 x 10¹⁴ h / Sec. = ν_N    

The fore explanation could verify in the practice, but this can not satisfy us; so, here will be given a theoretical explanation, that at the same time that coincide with the practice will support our models of atom with polygonal orbits and my model of the multimolecules, and my theory of electric current; beside many other things. In the model of refraction of light in the atmosphere (and in any medium); will consider that for an atom of the medium, to affect to one corpuscle of the refracted radiation, the deflection point or vertex of the affected electron, ought to be in front and in contact with the corpuscle of the radiation. For to attain this, the electron must move around the atom during a time of jump: tj = 1.12 x 10^—11 Sec. . The most intense light radiation that fulfill the fore condition is the violet one: ν = 7.5 x 10¹⁴ corpuscles / Sec.; 7.5 x 10¹⁴ / c =7.5 x 10¹⁴ / 3 x 10⁸ = 2.5 x 10⁶ Corps. / radiation (= nm…?);   t_m  =  t_j n_m = 1.12 x 10^—11 x 2.5 x 10⁶ = 2.8 x 10^—5 Sec. = time of multimolecule.

The resulted radiation will be a refracted one time, is:    ν_r = ν₄ – v_N = 7.5 x 10¹⁴ - 1.4336 x 10¹⁴ = 6.0664 x 10¹⁴ h / Sec. (blue color radiation); its refraction index is:   i_r = 7.5 / 6.0664 = 1.236.

After move the ray in the atmosphere during a time equal to: tm = 2,8 x 19^—5 Sec, in a distance approximate equal to:   d = c_r  t_m = (3 x 10⁸ / 1.236) 2.8 x 10^—5 = 6756 m; is produced other refringent effect and the ray gets a frequency:   ν c´ =6.0664 x 10¹⁴ – 1.4336 x 10¹⁴ = 4.5332 x 10¹⁴ (red color radiation).

It is consider that in the instant is produced the refraction in the affected radiation, the light diminishes its velocity; but after that, it gets light velocity, but with bigger (λ) and less (ν).

I have the idea that with mechanophysics is possible to find all the characteristics of the atoms. Many of them have been obtained by experiments, as the radii of them. In the data required in this theme, has been see that due to the atoms of the refringent substance are produced the refraction index in the radiations that penetrate into the refringent substance, here will be explained this, although for to obtain an exact value, it is not easy, because intervene some data that can have many variations due to many facts., In the models will be given here have been consider that the molecable electrons are those that produce the energy of the refraction effects in the incident radiations; for to understand better this problem, will consider a typical atom.

Next will be given a model of refringent substance, of course it ought to be a transparent one. We consider that for a body to be a stable one, all its atoms ought to produce their deflection effect at the same time. This also happens in other fields, for instance in the transfusion of blood; in the transplant of organs; in all bodies including the transparent and opaque ones; other way would be rejection effects in such bodies. Based in this will be made our model of refringent substance. If light reduces its velocity when enter to a refringent body is because its atoms affect the velocity of the incident radiation. We suppose that the main influence for this are the molecable electrons; their kinetic energy can reduce the velocity of the incident radiation; this energy is proportional to the square of their radii. The radial kinetic energy of the typical molecable electron is: K_^7 = 2.1747 x 10^—21 joule / t_v7 = 2.6296 x 10^—5 joule /Sec. = 3.2823 x 10¹² h /t_v7

Of course the fore energy affect the incident radiation, nevertheless, till here has not defined the time it acts. The fore energy acts in order the orbital velocity remains by indefinite time, because in every vertex is produced this radial deflection in every time of vertex: t_v7 = 8.27 x 10^—17 Sec. The fore energy acting N times, as will be seen afterward, can produce the energy that will refract the incident ray; all these in the case that the refringent body has the atoms (typical) with the values given before; but some atoms have radii bigger, or smaller than:  r_μ = 1 μ. Here have talked of substances formed by atoms, but there are other formed by molecules.

Till here have been defined the energies that produce the molecable electrons that affect the incident radiations. These energies K_^7 are produced every time of vertex t>₇, but has not been explained how acts in the corpuscles of the incident radiations. In other themes have been explained that the times of vertex are constant values for all the seven orbits of all the atoms, no matter they have different radii. In our model is consider that for each corpuscle of the incident radiation to be affected by E_^7´ energy, it coincide with the orbital electron that produces it; beside other conditions…. Taking in account the atoms (typical) with: r = 1 μ, distributed in an isometric way, and the incident ray with velocity: c = 3 x 10⁸ m / Sec (as if were not affected by the refringent body…). In this theoretical case, each corpuscle of the ray, for to advance from one atom to the adjacent one requires a time: tr = 2 r / c = 2 x 10^—10 / 3 x 10⁸ = 6.6667 x 10^—19 Sec. For coincide a corpuscle of light with an orbital electron in its time of vertex, the corpuscle ought to advance the following quantity of atoms:   n_>  = t_v7 / t_r = 8.27 x 10^—17 / 6.6667 x 10^—19 = 124 atoms.

As the molecable electrons of the atoms of the refringent substance have different values: r ≠ 1 μ, the values of n> can vary in a signify way. In a refringent substance formed by molecules the atoms that produce the refringent effect have other variations. Because all these, is not easy to determine the refringent index in this way; maybe there is a relation between the specific heat and the refraction index. In our model will be consider that if the refringent substance is formed by molecules; the atom of a molecule that is affected by the bigger quantity of molecable electrons, is that that contribute to the refraction of the incident radiation. The numerical examples will be given here fulfill the condition of an atom affected by 4 molecable electrons, except the diamond.                                                

First let see the water: H2O. The O has two mlecable electrons and is affected by two molecable electrons of the two H atoms. In an approximate way we suppose that its radius is: r ≈ (r_H + r_O) / 2 = (0.37 + 0.74) / 2 = 0.555 m. All light radiations get equal velocity in the refringent medium, but as they have different frequency ν when the radiation enter to such medium, its frequency diminishes in the same proportion than its velocity; the corpuscular separation does not change.. The light radiations have different energies h / Sec.) The violet color: W_v = c / λ_v = 3 x 10⁸ / 4 x 10^—7 = 7.5 x 10¹⁴ h /Sec. The red color   n_r = c / λ_r = 3 x 10⁸ / 7 x 10^—7 = 4.2857 x 10¹⁴ h / Sec.

Here will choose an average radiation ray:
W = (W_v W_r)^0.5 = (7.5 x 10¹⁴ x 4.2857 x 10¹⁴)^0.5 = 5.6695 x 10¹⁴ h / Sec.
Here will anticipate, applying some ideas we have not study well yet., as the structure of the liquids
 

, and of the molecule formed by more than two atoms. For the water will consider an (O) atom with two H ones, fixed to it perpendicularly, forming an L: But this molecule is not a move less, but has a double rotation movement, around atom (O) For produce such rotation it is required a minimum space that is that can be confined in a cube, that we will call: confined volume; the size of the fore mentioned cube has a length of 3 diameter of an atom, as will be seen forward. In the water can be consider the cubes of the confined volume distribute in an isometric way, in contact one with each other, so there is attraction between them with an energy equal to the cohesion there is in the water molecules. Here we have to remember that particles with the same sign do not reject, if they are not free to be oriented for this; on the contrary, can be attraction between them.

When were study the molecules, it was take in account that the atoms that form them ought to have the same diameter; because this, in an approximate way can have: r = 0.5(r_O + r_H) = 0.555 μ (see table of radii of atoms). It was say before that a water molecule is into the confined volume. This cube must have the density of water, given by the weight of the molecule of water, and also the required space for the double and perpendicular rotation was say before. The double rotation is produced simultaneously, and can be appreciated in detail considering rotation angles equal to: 360^o / 28; nevertheless and in order to simplify the explanations, here will consider the rotation angles equal to 90^o. In the annex figures is seen the confined volume corresponding to a rotating water molecule. Fig.(1) is a lateral view of the confined volume, indicating angular movements of 90^o with the numbers: 1, 2,. . .6. The fore points in the sphere correspond to the center of the sides of the cube of the confined volume. Fig.(2) is a superior view of such movements. Fig.(3) indicates the water molecule in the position: 4 – o – 1; point (vertex) 4 is the atom Ha; in (o) is atom (O); in 1 is atom Hb. Next will take in account the movement of the molecule, first the movement of atom Hb, starting from point 1 to 2, rotation axis 3-5; next from point 2 to 3, rotation axis 1-6; from point 3 to 4; and from 4 to 1. Rotation of atom Ha: from 4 to 5, rotation axis 1-6; from 5 to 6; from 6 to 3; from 3 to 4.

Next will take in account the rotation with 90^o angles, of the water molecule in form of L, considering the atom Ha the first term into the parenthesis; the atom O in the vertex (o); and the atom Hb the last term in the parenthesis. Starting rotation of the water molecule from: (4 – o – 1) (4 – o – 1); (5 – o – 2); (6 – o – 3); (3 – o – 4); (4 – o –1); and so on. . .

The weight of the molecule of water is: w = 18 x 1.674 x 10^—27 = 3.0132 x 10^—26 Kg. = 3.0132 x      10^—29 m³ of water = confine volume = (6 r)³ = 216 r³                                                                                                                     r = (3.0132 x 10^—29 / 216)^1/3  =  5.1863 x 10^—11 m;           φ = 1.0373 x 10^—10 m

The energy that have the molecable electrons that affect the O atom are those that produce the refringent effect in the incident ray, vary in a proportion equal to:   r_μ²  =  (r_μ / r)²  K_^7´ =  4 r_μ ² K_^7 =     4 (0.5186)² 3.2823 x 10¹² = 3.531 x 10¹² h / t_v7    By experiment has been determine the refraction index of water:   i_r = 1.33, so, the diffracted velocity of light in water is: cr = c / i_r = 3 x   10⁸ /1.33 = 2.2556 x 10⁸ m / Sec.   The incident radiation with c = 3 x 10⁸ m / Sec., gets the cr velocity since it enters to the periphery atoms of the water recipient; so the average velocity in such space is:  c_r´ = (c  c_r)^0.5  =  (3 x 10⁸ x 2.2556 x 10⁸)^0.5 = 2.6013 x 10⁸  m / Sec.

Will be determine the distance (d) that advance the incident radiation in order to be affected by the vertex effects:   d = c_r´ t_v7 = 2.6013 x 10⁸ x 8.27 x 10^—17 = 2.1513 x 10^—8 m The fore distance is equal to the following quantity of diameter of atoms:  n_>  =  d / 2 r = 2.1513 x 10^—8 / 1.0373 x 10^—10 = 207.4

The lost of kinetic energy of the incident radiation is: W_L = W (1 – 1/i_r) = 5.6695 x 10¹⁴ ( 1 – 1 / 1.33) = 1.4067 x 10¹⁴ h / Sec. Quantity of active atoms, acting in all the corpuscles of the radiation, required for produce the refringence of the incident ray:  N = W_L / K_^7´ = 1.4067 x 10¹⁴ / 3.531 x 10¹² = 39.84 atoms Distance travel by a radiation in order to gets refracted: D = n_> N φ = 207.4 x 39.84 x 1.0373 x 10^—10 = 8262.8 x 1.0373 x 10^—10 = 8.571 x 10^—7 m

It has been defined the quantity of time that an incident radiation pass by an atom in the instant it is producing a deflection effect. In a trajectory equal to (d). in water, the fore distance is equivalent to: n_> = 207.4 diameter of atoms. The time required for produce such refringence is:   t_r = N t_v7 = 39.84 x 8.27 x 10^—17 = 3.295 x 10^—15 Sec. The radiation into the refringent medium (the water) moves at an average velocity:  c_r´ = 2.6013 x 10⁸ m / Sec.; so the distance required by the radiation to gets regfracted is:   D = c_r´ t_r  = 2.6013 x 10⁸ x 3.295 x 10^—15 = 8.5713 x 10^—7 m.   Also is equal to:  D =     N d = 39.84 x 2.1513 x 10^—8. Alcohol is a liquid refringent medium, could be tried in a similar way than water.

In a refringent liquid substance as water was seen that the refringent molecules confined in their corresponding confine volume have capacity for produce a double rotation. In a solid refringent body, its molecules can not have the rotation movement; they are fixed in their confine volume, although do not occupy all its volume. As have say, all the atoms of a molecule get an equal diameter, in such way that an atom can be confines in a cube, whose sides are equal to the diameter of the atom; to this cubes we will call: atom cube. We can divide the confined volume by a whole number of atom cubes, and also all the refringent body, distributing them in an isometric way. Part of these cubes will be occupied by the molecules of the refringent body; so we have to define how are structured such molecules and how are they distributed in such body. The water molecules are formed as an L, in which the vertex of it is occupied by the atom (O), and the two branches are occupied by the two H atoms; this is logic because in an atom with two molecable electrons, they try to move in planes spaced at a maximum angle (90^o). For the quartz molecule will consider the Si atom in the vertex of the L and the two O atoms in the two branches of the L structure. It was defined the structure of the quartz molecule, next will be define how are distributed such molecules in the refringent body. We consider the two L in a vertical position, one is parallel to a lateral plane and the other parallel to a frontal plane of the refringent body; in such way that the Si atom of the first molecule is in contact with an

(O) atom of the other molecule; in this condition it is required to take in account a double confine volume, containing the two molecules; this double confine volume will have the following dimensions, expressed in diameters of atom: 3 x 2 x 2 = 12 atom cubes.

All said before was for to explain the conditions in which works the quartz atom; but as there are many refringent substances it is better to give the condition of all the refringent substances in a general way; forward will be done this for 2 or 3 atoms molecules: 1) Although can give different position to the refringent molecules, always it is possible to work with a single confine volume; for instance in the quartz molecules was obtained a double confine volume of 12 atom cubes with 2 molecules; this is equivalent to a single confine volume of 6 atom cubes with one molecule with 3 atom cubes That all the atom cubes are distributed in an isometric way in the refringent body For simplify the explanations will give a position to all the terms; so the refringent body has a horizontal plane formed by confine volumes. Parallel to it and spaced one confine volume are other horizontal planes similar to the fore one and covering all the refringent body

In a vertical way there will be lines formed with confine volumes; in few words we have an isometric distribution of atom cubes (that form the confine volumes) in the 3 dimension of the refringent body.

Let see the quartz molecule

Considering two L structures of this molecule, adjacent one to each other, as was specified before, seen one in frontal position and the other perpendicular we have a double confine volume; with atom cubes equal to to: 3 x 2 x2 = 12, with two molecules formed by 3 atom cubes each one; this is equivalent to a confine volume of 6 atom cubes, with one molecule. The confine volume is: V_c = 6  (2 r)³; the weight of the quartz molecule is: w = (28 + 2 x 16) 1.674 x 10^—27 = 1.0044 x 10^—25 . The density of quartz is: Δ = 2.5  to  2.8; its average one is:  Δ´ = 2.65.   With the fore values can be made the following proportion:    V_c : w :: 1 m³ : 2650 Kgs      V_c = 6 (2 r)³  =  48 r³ = w / 2650 = 1.0044 x 10^—25 / 2650 = 3.7902 x 10^--29 r = (3.7902 x 10^—29 / 48)^1/3 = 9.2429 x 10^--11

The active energy that has the (Si) atom of the molecule of quartz is: K_^7´ = 4  r_μ ² K_^7 = 4 (0.9243)² x 3.2823 x 10¹²  =  1.1216 x 10¹³ h / t_v7 The rafraction index of quartz is:   i_r = 1.46 The velocity of light into quartz is:   c_r = 3 x 10⁸ / 1.46 = 2.0548 x 10⁸ m / Sec. Average velocity of refringence:  c_r´ = (c c_r)^0.5 = (3 x 10⁸ x 2.0548 x 10⁸)^0.5 = 2.4828 x 10⁸ m / Sec. Dstance between active atoms of the molecules:   d = c_r´ t_v7 = 2.4828 x 10⁸ x 8,27 x 10^—17 = 2.0533 x 10^—8 m The fore distance expressed in diameter of atoms:   n_>7  =  d / 2 r = 2.0533 x 10^—8 / (2 x 9.2429 x 10^—11 = 111.07